Is there a technique for this?

[Guest]

What is the smallest number which must be added to 368 to make it exactly divisible by 27?

Is it just a case of performing a long division and seeing how much needs to be added to the remainder to make it up to 27 or is there a cleverer way?

[yoyo123]

not sure if it's any quicker, but I've just multiplied 27 by 10...270, then added mutiples of 27.

270 +54 = 324 (not enough)

another 27 =351 (not enough)

another 27 = 378 (10 over) so you need to add 10

[Guest]

I'd say division so you get 13 something = 368, so you do 14x 27 which gives you 378 so 10 is the number.

[Guessed!]

Not sure if this is worth the typing time but this method seems to work. It's a bit long-winded but at least it saves long division - providing the divisor isn't prime. Problem is explaining it

Most kids looking at the above example should instantly spot that 27 is the product of 9 and 3 - i.e. instant factorisation! Using the factors in any order they divide by first one (say 9) and then divide the integer of their answer by the other. They multiply their second remainder by their first divisor, add their first remainder and deduct it from the original divisor. Their answer will be what they have to add to the given number to reach the next multiple of the original divisor.

Here goes!

368 / 9 = 40 [r8]
40 / 3 = 13 [r1]
1 x 9 = 9 + 8 = 17
27 - 17 = 10

Here's another:

What needs to be added to 482 to make it divisible by 24?

24 = 6 x 4

482 / 6 = 80 [r2]
80 / 4 = 20 [r0] (Aha!!)
0 x 6 = 0 + 2 = 2
24 - 2 = 22

Test: 482 + 22 = 504 / 24 = 21

or the other way ...

482 / 4 = 120 [r2]
120 / 6 = 20 [r0]
0 x 4 = 0 + 2 = 2
24 - 2 = 22

Yup, it works!